Freezing point of urea solution is -0.6° C. How much urea ( m . wt .=60 g / mol ) will be required to dissolve in 3 kg water ? (kf=1.5° C kg mol -1)

Question

Freezing point of urea solution is -0.6° C. How much urea ( m . wt .=60 g / mol ) will be required to dissolve in 3 kg water ? (kf=1.5° C kg mol -1) (A) 24 g (B) 36 g (C) 60 g (D) 72 g

Tardigrade Admin · Accepted Answer

Δ Tf=(kf × 1000× w/M× W) w =(Δ Tf × M × W/kf × 1000) Δ Tf =0-(-0.6)=0.6° C M =60 g / mol kf =1.5° C kg mol -1 W =3 × 103 g w =(0.6 × 60 × 3 × 103/1.5 × 1000) =72 g

Q. Freezing point of urea solution is −0.6∘C. How much urea (m.wt.=60g/mol) will be required to dissolve in 3kg water ? (kf​=1.5∘Ckgmol−1)

24 g

36 g

60 g

72 g

ΔTf​=M×Wkf​×1000×w​ w=kf​×1000ΔTf​×M×W​ ΔTf​=0−(−0.6)=0.6∘C M=60g/mol kf​=1.5∘Ckgmol−1 W=3×103g w=1.5×10000.6×60×3×103​ =72g

Q. Freezing point of urea solution is $- 0. 6^{\circ} C$ . How much urea $(m . wt . = 60 g / m o l)$ will be required to dissolve in $3 k g$ water ?
$(k_{f} = 1. 5^{\circ} C k g m o l^{- 1})$